Let $f \in L^1(\mathbb{R}_+, \mathbb{R})$. It's Laplace transform is well defined for all $\Re(z) \geq 0$. Assume that: $$ \forall \Re(z) \geq 0,~ \widehat{f}(z) := \int_0^\infty{e^{-zt}f(t) dt} = G(z),$$ where $G$ is an Holomorphic function on $\Re(z) > -1$.
May I deduce that for all real number $\lambda < 1$, $t \mapsto f(t)e^{\lambda t} \in L^1(\mathbb{R}_+, \mathbb{R})$?
If not, do you have a counterexample? Thanks!
Not in general. For $\Re(s) > 0$ let $$f(t) = \cases{(-1)^{\textstyle\lfloor e^t \rfloor} \text{ if } t > 0\\ 0 \text{ otherwise}}$$ $$ F(s) = \int_0^\infty f(t)e^{-st}dt= \sum_{n=1}^\infty (-1)^n \int_n^{n+1} x^{-s-1}dx=\sum_{n=1}^\infty (-1)^n \frac{n^{-s}-(n+1)^{-s}}{s}$$ As $n^{-s}-(n+1)^{-s} = s n^{-s-1}+O(s(s+1)n^{-s-2})$ the latter series for $F(s)$ converges and is analytic for $\Re(s) > -1$.
But $f(t)e^{-\sigma t}$ is $L^1(\Bbb{R})$ only for $\sigma > 0$.
For $\Re(s) \in (-1,0)$ $$ F(s) = \int_{-\infty}^\infty (f(t)-1/2)e^{-st}dt$$ and the inverse Fourier transform of $F(\sigma+i\omega)$ in the sense of distributions gives $(f(t)-1/2)e^{-\sigma t}$